Visual SolutionPYQ 2024 · Jan Shift 2Tricky

Question

A company has two factories. Factory I produces 60% of total items and Factory II produces 40%. The defective rates are 2% and 3% respectively. An item is found defective. The probability it came from Factory II is:

(A)

\frac{21}{50}

(B)

\frac{1}{2}

(C)

\frac{2}{5}

(D)

\frac{29}{50}

Solution Path

P(F_1 \cap D) = 0.6 \times 0.02 = 0.012 = P(F_2 \cap D) = 0.4 \times 0.03

. By Bayes:

P(F_2|D) = 0.012/0.024 = 1/2

01Question Setup

1/4

Factory I: 60% production, 2% defective. Factory II: 40% production, 3% defective. Item is defective - find

P(\text{from Factory II})

Find

P(F_2 \mid D)

02Setup Probabilities

2/4

Joint probabilities:

P(F_1 \cap D) = 0.6 \times 0.02 = 0.012

and

P(F_2 \cap D) = 0.4 \times 0.03 = 0.012

. Both are equal!

P(F_1 \cap D) = P(F_2 \cap D) = 0.012

03Bayes TheoremKEY INSIGHT

3/4

P(F_2 \mid D) = \dfrac{P(F_2 \cap D)}{P(D)} = \dfrac{0.012}{0.012 + 0.012} = \dfrac{0.012}{0.024} = \dfrac{1}{2}

P(F_2 \mid D) = \dfrac{0.012}{0.024} = \dfrac{1}{2}

04Final Answer

4/4

Equal joint probabilities mean the defective item is equally likely from either factory.

P(F_2 \mid D) = \boxed{\dfrac{1}{2}}

- Answer (B)

Concepts from this question2 concepts unlocked

★

Bayes' Theorem

STANDARD

Reverse conditional probability: find P(cause|effect) from P(effect|cause) using prior probabilities.

P(A_i \mid B) = \frac{P(A_i) \cdot P(B \mid A_i)}{\sum_j P(A_j) \cdot P(B \mid A_j)}

JEE loves 'defective item from which factory' questions. Bayes flips the conditional -a pattern that feels unintuitive without practice.

Factory defectiveDisease testingMulti-source selection

Practice (12 Qs) →

Law of Total Probability

STANDARD

If events A₁, A₂, … partition the sample space, then P(B) = Σ P(Aᵢ) × P(B|Aᵢ).

P(B) = \sum_{i=1}^{n} P(A_i) \cdot P(B \mid A_i)

The denominator of Bayes' theorem. Also essential for any problem that says 'from one of several sources'.

Bayes denominatorMulti-urn problemsMixed populations

Practice (11 Qs) →

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